In "The Probabilistic Method" (4 ed.) by Noga Alon and Joel H. Spencer, on page $9$ following Proposition $1.3.1$, the authors have the following expression for $a,n,v\in \mathbb{N}; n,a\le v, b=v-a$
$$\frac{\binom{a}{n}+\binom{b}{n}}{\binom{v}{n}}\;.$$
They remark that "As $\binom{y}{n}$ is convex, this expression is minimized when $a=b$." Why is this true? Why does convexity immediately imply that it is minimized at the average?
Let $f(x)$ be a convex function, and suppose that $a<b$. Convexity of $f$ means that the line segment whose endpoints are $\langle a,f(a)\rangle$ and $\langle b,f(b)\rangle$ lies on or above the graph of the function between $a$ and $b$. If you want to express that algebraically, it says that for each $t\in[a,b]$,
$$f(t)\le\frac{f(b)-f(a)}{b-a}(t-a)+f(a)\;.$$
The point
$$\left\langle\frac{a+b}2,\frac{f(a)+f(b)}2\right\rangle$$
is the midpoint of the segment, and it lies at or above the point
$$\left\langle\frac{a+b}2,f\left(\frac{a+b}2\right)\right\rangle$$
on the graph of $f$. Thus,
$$\frac{f(a)+f(b)}2\ge f\left(\frac{a+b}2\right)\;,$$
and hence
$$f(a)+f(b)\ge 2f\left(\frac{a+b}2\right)\;.\tag{1}$$
In particular, if $a+b$ is fixed at $2c$ and we allow $a$ and $b$ to vary, the righthand side of $(1)$ is fixed at $2f(c)$, and $f(a)+f(b)$ is always at least that large, so $f(a)+f(b)$ is minimized when $a=b=c$.